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Hi I'd like to know if the following proof of Vandermonde's Identity is correct (is really easy):

Let $m,n,r$ be natural numbers such that $r\le \min \{m,n\}$. The Vandermonde's Identity gives us that $$ \binom{m+n}r = \sum_{k=0}^r \binom mk \binom n{r-k} $$

$\bf{Solution}$ Let $\gamma$ be the unit circle $\lvert z\lvert =1$. Now since $${m \choose r}= \frac{1}{2\pi i}\int _{\gamma}\frac{(1+z)^m}{z^{r+1}} dz$$

Then

\begin{align} 2\pi i \binom{m+n}r= \int _{\gamma}\frac{(1+z)^{m+n}}{z^{r+1}} dz &= \int _{\gamma}\frac{(1+z)^{n}}{z^{r+1}}\sum_{k=0}^m {m \choose k}z^k \,dz\\ &= \sum_{k=0}^m { m\choose k}\int_\gamma \frac{(1+z)^n}{z^{r-k+1}} dz \\ &= \sum_{k=0}^m 2\pi i { m\choose k} {n\choose r-k} \end{align}

For all $k>r$, we set ${n\choose r-k}=0$, i.e., the left-hand side is zero and imply the desired identity

$$ \binom{m+n}r = \sum_{k=0}^r \binom mk \binom n{r-k} $$

Thanks :)

Jose Antonio
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  • @gt6989b Do you think is correct? – Jose Antonio Sep 02 '15 at 03:18
  • yes, i think this is right – gt6989b Sep 02 '15 at 03:46
  • @gt6989b I like it because is incredible easy. Even more than the usual approach using generating functions. Thanks :) – Jose Antonio Sep 02 '15 at 03:59
  • @JoseAntonio Instead of using generating functions, I think an easy combinatoric proof is also possible. – PSPACEhard Sep 02 '15 at 05:29
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    Nice use of complex calculus. A combinatorial proof : Let A have m members and let B have n members with $ A\cap B = \phi$. For k from 0 to r , compute the number of ways to choose a set with r members with k of them in A and r-k of them in B. Now sum over k. – DanielWainfleet Sep 02 '15 at 08:16
  • @user254665 Thank you; a similar exercise is in Conway's Complex Analysis and when I was working with this identity yesterday I had the idea, why no put the contour integral to show it. Could you explain a little bit more this approach, please? I'm not very good using a combinatorial approach. – Jose Antonio Sep 02 '15 at 14:02
  • @user254665 The only point where I'm not completely sure is to set ${n, \choose k} \equiv 0$ when $k<0$ – Jose Antonio Sep 02 '15 at 14:04

1 Answers1

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To keep the question off the unanswered list. Yes, your approach is correct.

See Complex Analysis proof of multinomial expression for something in the same spirit.

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